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Red-Black Trees

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Red-Black Trees

DefinitionsandBottom-Up Insertion

- Definition: A red-black tree is a binary search tree in which:
- Every node is colored either Red or Black.
- Each NULL pointer is considered to be a Black “node”.
- If a node is Red, then both of its children are Black.
- Every path from a node to a NULL contains the same number of Black nodes.
- By convention, the root is Black

- Definition: The black-height of a node, X, in a red-black tree is the number of Black nodes on any path to a NULL, not counting X.

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X

A Red-Black Tree with NULLs shown

Black-Height of the tree (the root) = 3Black-Height of node “X” = 2

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A Red-Black Tree with

Black-Height = 3

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X

Black Height of the tree?

Black Height of X?

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Theorem 1 – Any red-black tree with root x, has n ≥ 2bh(x) – 1 nodes, where bh(x) is the black height of node x.

Proof: by induction on height of x.

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Theorem 2 – In a red-black tree, at least half the nodes on any path from the root to a NULL must be Black.

Proof – If there is a Red node on the path, there must be a corresponding Black node.

Algebraically this theorem means

bh( x ) ≥ h/2

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Theorem 3 – In a red-black tree, no path from any node, X, to a NULL is more than twice as long as any other path from X to any other NULL.

Proof: By definition, every path from a node to any NULL contains the same number of Black nodes. By Theorem 2, a least ½ the nodes on any such path are Black. Therefore, there can no more than twice as many nodes on any path from X to a NULL as on any other path. Therefore the length of every path is no more than twice as long as any other path.

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Theorem 4 –

A red-black tree with n nodes has height h ≤ 2 lg(n + 1).

Proof: Let h be the height of the red-black tree with root x. By Theorem 2,

bh(x) ≥ h/2

From Theorem 1, n ≥ 2bh(x) - 1

Therefore n ≥ 2 h/2 – 1

n + 1 ≥ 2h/2

lg(n + 1) ≥ h/2

2lg(n + 1) ≥ h

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- Insert node as usual in BST
- Color the node Red
- What Red-Black property may be violated?
- Every node is Red or Black?
- NULLs are Black?
- If node is Red, both children must be Black?
- Every path from node to descendant NULL must contain the same number of Blacks?

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- Insert node; Color it Red; X is pointer to it
- Cases
0: X is the root -- color it Black

1: Both parent and uncle are Red -- color parent and uncle Black, color grandparent Red. Point X to grandparent and check new situation.

2 (zig-zag): Parent is Red, but uncle is Black. X and its parent are opposite type children -- color grandparent Red, color X Black, rotate left(right) on parent, rotate right(left) on grandparent

3 (zig-zig): Parent is Red, but uncle is Black. X and its parent are both left (right) children -- color parent Black, color grandparent Red, rotate right(left) on grandparent

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G

X

P

U

G

X

P

U

Case 1 – U is Red

Just Recolor and move up

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G

P

U

X

X

S

P

G

Case 2 – Zig-Zag

Double Rotate X around P; X around G

Recolor G and X

S

U

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G

P

U

S

P

X

X

G

Case 3 – Zig-Zig

Single Rotate P around G

Recolor P and G

U

S

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- O(lg n) to descend to insertion point
- O(1) to do insertion
- O(lg n) to ascend and readjust == worst case only for case 1
- Total: O(log n)

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An alternative to this “bottom-up” insertion is “top-down” insertion.

Top-down is iterative. It moves down the tree, “fixing” things as it goes.

What is the objective of top-down’s “fixes”?

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11

Insert 4 into this R-B Tree

14

2

15

1

7

5

8

Red node

Black node

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Insert the values 2, 1, 4, 5, 9, 3, 6, 7 into an initially empty Red-Black Tree

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